Sum of a series $\frac {1}{n^2 - m^2}$ m and n odd, $m \ne n$

Question

I was working on a physics problem, where I encountered the following summation problem: $$ \sum_{m = 1}^\infty \frac{1}{n^2 - m^2}$$ where m doesn't equal n, and both are odd. n is a fixed constant

We could alternatively write: $$ \sum_{j = 1}^\infty \frac{1}{(2i - 1)^2 - (2j - 1)^2}$$ where i doesn't equal j. I used a numerical solver to solve for a few values of n, and the solution seems to be $-1/(4n^2)$ which fits with the problem, but I'd like to find a rigorous solution.

My attempts so far have involved simplifying the term in the sum in various was to see if I could find a way solve the problem, but I've had no luck.

Ex. The second equation can be simplified to $$\sum_{j=1}^\infty \frac {1} {4(i + j - 1)(i - j)} $$ but I'm not sure where to go from here.

I've also tried splitting the problem into two sections, one where m < n, and one where m > n, but again, I'm not sure how to continue.

Answer 1

Assume $N>i\geq1$. Starting by a partial fraction decomposition, one may write $$ \begin{align} &\sum_{j=1,\,j\neq i}^N\frac1{(2i-1)^2-(2j-1)^2} \\\\&=\frac1{4(2i-1)}\sum_{j=1,\,j\neq i}^N\left(\frac1{j+i-1}-\frac1{j-i}\right) \\\\&=\frac1{4(2i-1)}\left(\sum_{j=1}^N\frac1{j+i-1}-\frac1{2i-1}-\sum_{j=1,\,j\neq i}^N\frac1{j-i}\right) \\\\&=\frac1{4(2i-1)}\left(\sum_{\ell=i}^{N+i-1}\frac1{\ell}-\frac1{2i-1}-\sum_{j=1}^{i-1}\frac1{j-i}-\sum_{j=i+1}^N\frac1{j-i}\right) \\\\&=\frac1{4(2i-1)}\left(H_{N+i-1}-H_{i-1}-\frac1{2i-1}+H_{i-1}-H_{N-i}\right) \\\\&=-\frac1{4(2i-1)^2}+\frac1{4(2i-1)}\left(H_{N+i-1}-H_{N-i}\right) \end{align} $$ then, as $N \to \infty$,

$$ \sum_{j=1,\,j\neq i}^\infty\frac1{(2i-1)^2-(2j-1)^2}=-\frac1{4(2i-1)^2} $$

where $H_N$ is the harmonic number and where we have used $$ H_{N+i-1}-H_{N-i}=O\left(\frac1{N}\right) $$ as $N \to \infty$.

Answer 2

The case of all integers is $$ \sum_{m=1}^{n-1}\frac{1}{n^2-m^2}+\sum_{m=n+1}^{\infty}\frac{1}{n^2-m^2} = \frac{-3}{4n^2} $$ and the "odd" case is similar $$ \sum_{k=1}^{j-1}\frac{1}{(2j-1)^2-(2k-1)^2} +\sum_{k=j+1}^\infty\frac{1}{(2j-1)^2-(2k-1)^2} = \frac{-1}{4(2j-1)^2} $$ These may be computed with information about the "digamma" function $\psi(z) = \Gamma'(z)/\Gamma(z)$, namely: $$ \sum_{k=u}^v \frac{1}{a+k} = \psi(a+v+1)-\psi(a+u), \qquad\text{where }u<v\text{ integers and } a\text{ complex} $$ and $$ \psi(z) = \log z -\frac{1}{2z} + O\left(\frac{1}{z^2}\right)\text{ as } z \to +\infty $$

Answer 3

If we start with: $$ \cos\left(\frac{\pi x}{2}\right) = \prod_{k\geq 0}\left(1-\frac{x^2}{(2k+1)^2}\right) $$ we get, by considering the logarithmic derivative: $$ \frac{\pi}{4x}\cdot\tan\left(\frac{\pi x}{2}\right) = \sum_{k\geq 0}\frac{1}{(1+2k)^2-x^2}$$ that is a meromorphic function with a regular point in the origin and simple poles at the odd integers: the residue at $x=\pm(2K+1)$ is just $\mp\frac{1}{4K+2}$. We also have: $$\begin{eqnarray*} \sum_{\substack{k\geq 0 \\ k\neq K}}\frac{1}{(2k+1)^2-(2K+1)^2} &=& \lim_{x\to(2K+1)}\left(\frac{\pi}{4x}\cdot\tan\left(\frac{\pi x}{2}\right)-\frac{1}{(2K+1)^2-x^2}\right)\\&=&\lim_{x\to(2K+1)}\left(\frac{\pi}{4x}\cdot\tan\left(\frac{\pi x}{2}\right)-\frac{\frac{1}{4K+2}}{(2K+1)-x}-\frac{\frac{1}{4K+2}}{(2K+1)+x}\right)\\&=&\color{red}{-\frac{1}{(4K+2)^2}}\end{eqnarray*}$$ since we know in advance what the singular part at $x=2K+1$ is.